Tensor Operator
Tensor times matrix: the k-mode product
The k k k-mode product of a tensor X ∈ R I 1 × I 2 × … × I N \boldsymbol{X} \in \mathbb{R}^{\boldsymbol{I}_{1} \times \boldsymbol{I}_{2} \times \ldots \times \boldsymbol{I}_{N}} X∈RI1×I2×…×IN with a matrix A ∈ R J × I k A \in \mathbb{R}^{J \times I_{k}} A∈RJ×Ik is written as
Y = X × k A Y=X \times_{k} A Y=X×kA
The resulting tensor Y Y Y is of size I 1 × … × I k − 1 × J × I k + 1 × … × I N I_{1} \times \ldots \times I_{k-1} \times J \times I_{k+1} \times \ldots \times I_{N} I1×…×Ik−1×J×Ik+1×…×IN, and contains the elements
y i 1 ⋯ i k − 1 j i k + 1 ⋯ i N = ∑ i k = 1 I k x i 1 i 2 ⋯ i N a j i k y_{i_{1} \cdots i_{k-1} j i_{k+1} \cdots i_{N}}=\sum_{i_{k}=1}^{I_{k}} x_{i_{1} i_{2} \cdots i_{N}} a_{j i_{k}} yi1⋯ik−1jik+1⋯iN=ik=1∑Ikxi1i2⋯iNajik
A few important facts about the k-mode product.
– X × m A × n B = X × n B × m A X \times_m A \times_n B = X \times_n B \times_m A X×mA×nB=X×nB×mA if n ≠ m n \neq m n=m
– but X × n A × n B = X × n ( B A ) X \times_n A \times_n B = X \times_n (BA) X×nA×nB=X×n(BA) (in general ≠ X × n B × n A \neq X \times_n B \times_n A =X×nB×nA)
Tucker Composition
For a 3rd-order tensor T ∈ F d 1 × d 2 × d 3 T \in F^{d_{1} \times d_{2} \times d_{3}} T∈Fd1×d2×d3, where F F F is either R \mathbb{R} R or C \mathbb{C} C, ‘’‘Tucker Decomposition’’’ can be denoted as follows,
T = T × 1 U ( 1 ) × 2 U ( 2 ) × 3 U ( 3 ) T = \mathcal{T} \times_{1} U^{(1)} \times_{2} U^{(2)} \times_{3} U^{(3)} T=T×1U(1)×2U(2)×3U(3)
where T ∈ F d 1 × d 2 × d 3 \mathcal{T} \in F^{d_{1} \times d_{2} \times d_{3}} T∈Fd1×d2×d3 is the ‘‘core tensor’’, a 3rd-order tensor that contains the 1-mode, 2-mode and 3-mode singular values of T T T, which are defined as the ''Frobenius norm" of the 1-mode, 2-mode and 3-mode slices of tensor T \mathcal{T} T respectively. U ( 1 ) , U ( 2 ) , U ( 3 ) U^{(1)}, U^{(2)}, U^{(3)} U(1),U(2),U(3) are unitary matrices in F d 1 × d 1 , F d 2 × d 2 , F d 3 × d 3 F^{d_{1} \times d_{1}}, F^{d_{2} \times d_{2}}, F^{d_{3} \times d_{3}} Fd1×d1,Fd2×d2,Fd3×d3 respectively. Note that T \mathcal{T} T might be much smaller than the original tensor T T T if we accept an approximation instead of an exact equality. The CP decomposition can be seen as a special case of the Tucker decomposition, where the core tensor T \mathcal{T} T is constrained to be superdiagonal.
The ‘‘j’’-mode product (’‘j’’ = 1, 2, 3) of T \mathcal{T} T by U ( j ) U^{(j)} U(j) is denoted as T × U ( j ) \mathcal{T} \times U^{(j)} T×U(j) with entries as
( T × 1 U ( 1 ) ) ( d 1 , d 2 , d 3 ) = ∑ i 1 = 1 d 1 T ( i 1 , d 2 , d 3 ) U ( 1 ) ( d 1 , i 1 ) ( T × 2 U ( 2 ) ) ( d 1 , d 2 , d 3 ) = ∑ i 2 = 1 d 2 T ( d 1 , i 2 , d 3 ) U ( 2 ) ( d 2 , i 2 ) ( T × 3 U ( 3 ) ) ( d 1 , d 2 , d 3 ) = ∑ i 3 = 1 d 3 T ( d 1 , d 2 , i 3 ) U ( 3 ) ( d 3 , i 3 ) (\mathcal{T} \times_{1} U^{(1)})(d_{1}, d_{2}, d_{3}) = \sum_{i_{1}=1}^{d_{1}} \mathcal{T}(i_{1}, d_{2}, d_{3})U^{(1)}(d_{1}, i_{1})\\ (\mathcal{T} \times_{2} U^{(2)})(d_{1}, d_{2}, d_{3}) = \sum_{i_{2}=1}^{d_{2}} \mathcal{T}(d_{1}, i_{2}, d_{3})U^{(2)}(d_{2}, i_{2}) \\ (\mathcal{T} \times_{3} U^{(3)})(d_{1}, d_{2}, d_{3}) = \sum_{i_{3}=1}^{d_{3}} \mathcal{T}(d_{1}, d_{2}, i_{3})U^{(3)}(d_{3}, i_{3}) (T×1U(1))(d1,d2,d3)=i1=1∑d1T(i1,d2,d3)U(1)(d1,i1)(T×2U(2))(d1,d2,d3)=i2=1∑d2T(d1,i2,d3)U(2)(d2,i2)(T×3U(3))(d1,d2,d3)=i3=1∑d3T(d1,d2,i3)U(3)(d3,i3)